Riemann Integration Mastery Workshop

From Individual Integrals to Limits of Functions

The preceding tutorials developed the Riemann integral for individual functions and established tools for estimating it. A natural next question is what happens when a function is described as a limit of simpler functions. Can we prove that the limit is integrable by studying its approximations? When may we compute its integral by first integrating the approximations and then taking a limit?

Uniform convergence provides a reliable answer on a closed bounded interval. It ensures that all points are approximated well at once, so an estimate on the function values becomes an estimate on the upper and lower sums and, ultimately, on the integrals. Pointwise convergence alone does not provide this control.

Throughout, let \(a<b\), and consider bounded functions on \([a,b]\). Recall the Darboux Criterion: a bounded function is Riemann integrable if, for every positive tolerance, some partition has upper sum minus lower sum smaller than that tolerance. We will use this criterion, along with the Uniform Approximation Bound for Integrals, rather than re-establishing earlier results about individual integrals.

Uniform Limits Preserve Riemann Integrability

Definition (Uniform Convergence): A sequence of functions \(f_n:[a,b]\to\mathbb{R}\) converges uniformly to \(f:[a,b]\to\mathbb{R}\) if for every \(\varepsilon>0\), there is an integer \(N\) such that

|f_n(x)-f(x)|<\varepsilon \quad\text{for every }n\geq N\text{ and every }x\in[a,b].

The same index \(N\) works at every point of the interval.

Theorem (Uniform Limits and Riemann Integration): Suppose each \(f_n:[a,b]\to\mathbb{R}\) is Riemann integrable and \(f_n\) converges uniformly to \(f\). Then \(f\) is Riemann integrable, and

\lim_{n\to\infty}\int_a^b f_n(x)\,dx = \int_a^b f(x)\,dx.

Proof. First, \(f\) is bounded. Choose an index \(n_0\) such that \(|f(x)-f_{n_0}(x)|<1\) for every \(x\in[a,b]\). Since \(f_{n_0}\) is Riemann integrable, it is bounded; choose \(B\geq0\) such that \(|f_{n_0}(x)|\leq B\) throughout the interval. Then

|f(x)|\leq |f_{n_0}(x)|+|f(x)-f_{n_0}(x)|<B+1.

Thus the Darboux Criterion applies to \(f\). Let \(\varepsilon>0\). Uniform convergence gives an index \(n\) such that

|f(x)-f_n(x)|<\delta \quad\text{for every }x\in[a,b], \qquad \delta=\frac{\varepsilon}{8(b-a)}.

For each subinterval of any partition \(P\), the supremum of \(f\) is at most the supremum of \(f_n\) plus \(\delta\), and the infimum of \(f\) is at least the infimum of \(f_n\) minus \(\delta\). Multiplying these bounds by the subinterval lengths and adding gives

U(f,P)\leq U(f_n,P)+\delta(b-a), \qquad L(f,P)\geq L(f_n,P)-\delta(b-a).

Because \(f_n\) is Riemann integrable, the Darboux Criterion gives a partition \(P\) such that \(U(f_n,P)-L(f_n,P)<\varepsilon/2\). For this partition,

U(f,P)-L(f,P) \leq U(f_n,P)-L(f_n,P)+2\delta(b-a) <\frac{\varepsilon}{2}+\frac{\varepsilon}{4} <\varepsilon.

The Darboux Criterion now proves that \(f\) is Riemann integrable. Finally, the Uniform Approximation Bound for Integrals, applied to \(f_n\) and \(f\), gives

\left|\int_a^b f_n(x)\,dx-\int_a^b f(x)\,dx\right| \leq \delta(b-a)

whenever \(n\) is sufficiently large for the chosen \(\delta\). Given any desired positive error, uniform convergence lets us choose \(\delta\) small enough that this upper bound is below that error. Therefore the integrals converge to the integral of \(f\). \(\square\)

The proof has two distinct jobs. The Darboux-gap estimate establishes that the limit function is integrable; only after that has been shown can we compare its integral with the integrals of the approximating functions. In particular, the theorem does not assume that the limit function is integrable in advance.

Worked Example: Step Functions Approximating the Identity

For each positive integer \(n\), define \(s_n:[0,1]\to\mathbb{R}\) by \(s_n(x)=k/n\) when \(k/n\leq x<(k+1)/n\), for \(k=0,\ldots,n-1\), and set \(s_n(1)=1\). Each \(s_n\) is a step function, hence Riemann integrable. On each half-open subinterval, \(0\leq x-s_n(x)<1/n\); at \(x=1\), the difference is zero. Thus \(s_n\) converges uniformly to \(f(x)=x\).

The integral of each step function is the sum of its constant values times the subinterval lengths. The value at the single endpoint \(1\) does not affect its integral:

\int_0^1 s_n(x)\,dx =\sum_{k=0}^{n-1}\frac{k}{n}\frac{1}{n} =\frac{1}{n^2}\frac{(n-1)n}{2} =\frac{n-1}{2n}.

As \(n\to\infty\), this tends to \(1/2\). The uniform-limit theorem ensures that the limit function is integrable and that its integral is this limit; indeed, \(\int_0^1 x\,dx=1/2\). The example illustrates how an integral can be obtained from simple step approximations.

Worked Example: A Uniformly Small Oscillatory Perturbation

On \([0,1]\), let \(f_n(x)=x+\sin(nx)/n\). Since \(|\sin(nx)|\leq1\),

|f_n(x)-x|=\frac{|\sin(nx)|}{n}\leq\frac{1}{n} \quad\text{for every }x\in[0,1].

Consequently, \(f_n\) converges uniformly to \(f(x)=x\), and every \(f_n\) is continuous and therefore Riemann integrable. Direct calculation gives

\int_0^1 f_n(x)\,dx =\frac{1}{2}+\frac{1}{n}\int_0^1\sin(nx)\,dx =\frac{1}{2}+\frac{1-\cos(n)}{n^2}.

Here the antiderivative \(-\cos(nx)/n\) gives \(\int_0^1\sin(nx)\,dx=(1-\cos(n))/n\), which verifies the displayed expression. Since \(0\leq1-\cos(n)\leq2\), the final term tends to zero. The limiting integral is therefore \(\int_0^1x\,dx=1/2\), as the theorem predicts.

Integrating a Uniformly Convergent Series

A series of functions is handled by looking at its partial sums. If the terms are Riemann integrable and the partial sums converge uniformly, the theorem above applies directly. This gives a rigorous basis for integrating a series term by term under a uniform-convergence hypothesis.

Theorem (Term-by-Term Integration under Uniform Convergence): Suppose \(u_k:[a,b]\to\mathbb{R}\) is Riemann integrable for every nonnegative integer \(k\), and the series \(\sum_{k=0}^{\infty}u_k(x)\) converges uniformly on \([a,b]\) to \(S(x)\). Then \(S\) is Riemann integrable, the numerical series of integrals converges, and

\int_a^b S(x)\,dx = \sum_{k=0}^{\infty}\int_a^b u_k(x)\,dx.

Proof. Define the partial sums \(S_N(x)=\sum_{k=0}^{N}u_k(x)\). By linearity for finite sums, each \(S_N\) is Riemann integrable and

\int_a^b S_N(x)\,dx = \sum_{k=0}^{N}\int_a^b u_k(x)\,dx.

By hypothesis, \(S_N\) converges uniformly to \(S\). The Uniform Limits and Riemann Integration Theorem shows that \(S\) is Riemann integrable and that \(\int_a^b S_N(x)\,dx\) converges to \(\int_a^b S(x)\,dx\). The displayed finite-sum identity identifies each partial sum of the numerical series on the right with \(\int_a^b S_N(x)\,dx\). Taking the limit of these partial sums proves both convergence of the numerical series and the claimed equality. \(\square\)

Worked Example: Integrating a Geometric Series Uniformly

On \([0,1]\), consider the series \(\sum_{k=0}^{\infty}(x/2)^k\). Its partial sums are continuous. For every \(N\geq0\) and every \(x\in[0,1]\), the geometric-series remainder satisfies

\left|\sum_{k=0}^{\infty}\left(\frac{x}{2}\right)^k -\sum_{k=0}^{N}\left(\frac{x}{2}\right)^k\right| = \frac{(x/2)^{N+1}}{1-x/2} \leq 2^{-N}.

The bound tends to zero independently of \(x\), so the series converges uniformly. Its sum is \(S(x)=1/(1-x/2)\). Term-by-term integration is therefore justified:

\int_0^1\frac{1}{1-x/2}\,dx = \sum_{k=0}^{\infty}\int_0^1\left(\frac{x}{2}\right)^k\,dx = \sum_{k=0}^{\infty}\frac{1}{2^k(k+1)}.

For the integral on the left, substitute \(u=1-x/2\), so \(dx=-2\,du\). The endpoints \(x=0\) and \(x=1\) correspond to \(u=1\) and \(u=1/2\), respectively, giving \(2\int_{1/2}^{1}du/u=2\ln 2\). Thus the series of integrals also sums to \(2\ln 2\). The uniform remainder estimate, not just the pointwise geometric-series formula, is what licenses the interchange.

Why Pointwise Convergence Is Not Enough

Uniform convergence controls the largest error anywhere on the interval. In pointwise convergence, the index needed for a given accuracy may depend on the point, and substantial contributions to an integral can persist on intervals that change with the index. The next example shows that even integrable functions converging pointwise to an integrable function need not have integrals converging to its integral.

Worked Example: A Pointwise Limit with the Wrong Integral Limit

For each positive integer \(n\), define \(h_n:[0,1]\to\mathbb{R}\) by \(h_n(x)=n\) when \(0<x<1/n\), and \(h_n(x)=0\) otherwise. These are step functions, so each is Riemann integrable. At \(x=0\), every \(h_n(0)=0\). For any fixed \(x>0\), once \(n\geq1/x\), the point \(x\) is not in \((0,1/n)\), so \(h_n(x)=0\). Hence \(h_n(x)\to0\) for every \(x\in[0,1]\).

Nevertheless, the integral of each function is

\int_0^1 h_n(x)\,dx=n\left(\frac{1}{n}-0\right)=1.

The pointwise limit is the zero function, whose integral is zero. Therefore the integrals do not converge to the integral of the pointwise limit. The convergence is not uniform: for each \(n\), points in \((0,1/n)\) have error \(n\), which does not become uniformly small.

This example does not contradict the theorem: its essential hypothesis, uniform convergence, fails. A useful proof check is to identify the exact estimate that makes a limiting argument work. Here that estimate is \(|f_n(x)-f(x)|<\delta\) for every \(x\), with one index working across the entire interval. Without it, a small pointwise error at each fixed point need not control the total contribution to an integral.

A Practical Proof Strategy

When a function or series is presented through approximations, organize the argument around the following questions:

Check the approximants.
Verify that each function being integrated is Riemann integrable. Continuous functions and step functions are common choices already covered in this course.

Establish uniform control.
Find a bound on the approximation error that holds for every point of the interval and tends to zero with the index.

Apply the uniform-limit theorem.
This proves integrability of the limit and convergence of the integrals. For a series, apply it to the partial sums and then use finite linearity.

Compute only after justifying the passage.
Evaluate the approximating integrals and take their limit; do not infer an interchange merely from pointwise convergence.

The central technique is a uniform error estimate. It transforms information about function values into control of Darboux sums and integral values. That makes uniform approximation a practical tool for proving integrability, evaluating limits of integrals, and integrating function series—not merely a condition attached to an abstract theorem.

Check Your Understanding

Use the estimates and hypotheses in this tutorial to answer the following questions.

In the proof that a uniform limit is Riemann integrable, why is boundedness of the limit established before applying the Darboux Criterion?
If \(|f_n(x)-f(x)|\leq 1/n\) on an interval \([a,b]\), what upper bound does the Uniform Approximation Bound give for the difference between their integrals?
Why does uniform convergence of the partial sums of a series of Riemann integrable functions allow term-by-term integration?
For the sequence \(h_n\) in the final worked example, what is its pointwise limit, and what is \(\int_0^1 h_n(x)\,dx\) for each \(n\)?
What feature of uniform convergence is missing from pointwise convergence and is needed to control the integral error across the whole interval?